Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 NonTerminationProof (⇔, 244 ms)
↳2 NO

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a(x1) → b(x1)
a(b(x1)) → b(b(a(c(c(a(x1))))))
c(b(x1)) → x1

Q is empty.

(1) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite.

Found the self-embedding DerivationStructure:
a b b b → b b a b b b c c b c c b

a b b b → b b a b b b c c b c c b
by OverlapClosure OC 3
a b b b → b b a a b c c b
by OverlapClosure OC 3
a b b b → b b a a c b b c c b
by OverlapClosure OC 3
a b b b → b b a a c c b b b c c b
by OverlapClosure OC 2
a b b → b b a a c c a
by OverlapClosure OC 3
a b b → b b a c b a c c a
by OverlapClosure OC 3
a b b → b b a c c b b a c c a
by OverlapClosure OC 2
a b → b b a c c a
by original rule (OC 1)
a b → b b a c c a
by original rule (OC 1)
c b →
by original rule (OC 1)
c b →
by original rule (OC 1)
a b → b b b c c b
by OverlapClosure OC 3
a b → b b a c c b
by OverlapClosure OC 2
a b → b b a c c a
by original rule (OC 1)
a → b
by original rule (OC 1)
a → b
by original rule (OC 1)
c b →
by original rule (OC 1)
c b →
by original rule (OC 1)
a b → b b b c c b
by OverlapClosure OC 3
a b → b b a c c b
by OverlapClosure OC 2
a b → b b a c c a
by original rule (OC 1)
a → b
by original rule (OC 1)
a → b
by original rule (OC 1)

(2) NO